x(x + y) – 5x – 5y

= x( x+ y) – ( 5x + 5y)

= x(x + y) – 5(x + y) = (x + y)(x – 5)

\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)

\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x+y-5\right)\left(x-y\right)\)

\(x^2-y^2-5x+5y=\left(x+y\right)\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)

y(x-y)-5x+5y

=y(x-y)-(5x-5y)

=y(x-y)-5(x-y)

=(x-y)(y-5)

\(y\cdot\left(x-y\right)-5x+5y=y\cdot\left(x-y\right)-5\cdot\left(x-y\right)=\left(x-y\right)\cdot\left(y-5\right)\)

tích mình với

ai tích mình 

mình tích lại

thanks nhiều

\(=\left(x-y\right)\left(x+y\right)+5\left(x-y\right)\)

\(=\left(x+y+5\right)\left(x-y\right)\)

Học tốt

Ta có : x² - y² - 5x - 5y

= ( x² - y² ) - ( 5x + 5y )

= ( x - y )( x + y ) - 5( x + y )

= ( x + y )( x - y - 5 )

cảm ơn bạn moon nhé bạn đã giúp mình rất rất nhiều ( nick trên cx là cảu mình )

a) \(5x-20y\)

\(=5\left(x-4y\right)\)

b) \(5x\left(x-1\right)-3x\left(x-1\right)\)

\(=2x\left(x-1\right)\) 

c) \(x\left(x+y\right)-5x-5y\)

\(=x\left(x+y\right)-5\left(x+y\right)\)

\(=\left(x-5\right)\left(x+y\right)\)

\(x\left(x+y\right)-5x-5y\)

\(=x\left(x+y\right)-5\left(x+y\right)\)

\(=\left(x+y\right)\left(x-5\right)\)

x(x+y)-5x-5y

= x.x+x.y-5x-5y

= x^2+x.y-5x-5y

= x.(x-5)-y.(x-5)

= (x-y).(x-5)

Đúng thì k nhak ^3^"""

Sửa lại 2 dòng cuối ( gõ nhầm dấu mất tiêu =_=//)

.....

= x(x-5)+y(x-5)

=(x+y).(x-5)

\(5x^2+3\left(x+y\right)-5y^2=5\left(x+y\right)\left(x-y\right)+3\left(x+y\right)=\left(x+y\right)\left[5\left(x-y\right)+3\right]\)

a) = (x - 1)(5x - 3)

b) = x(x + y) - 5(x + y)

= (x + y)(x - 5)

c) = x(x - y) - y(x - y)

= (x - y)^2

d) = 2(x² + 2x + 1 - y²)

= 2[(x + 1)² - y²]

= 2(x - y + 1)(x + y + 1)